Question

if 2a - (2/a) + 1 = 0 then find out a^3 - (1/a^3) +2 = ?​

Answer 1

ANSWER :–

$\\ \longrightarrow{ \red { \boxed{ \bold{ {a}^{ 3} - \frac{1}{ {a}^{3} } + 2 = \frac{3}{8} }}}}$

EXPLANATION :–

GIVEN :–

• ${ \bold{2a - \dfrac{2}{a} + 1 = 0}}$

TO FIND :–

• Value of $\: \: { \bold{a^{3} - \dfrac{1}{a^{3} } + 2 = ? }} \: \:$

SOLUTION :–

$\\ \implies{ \bold{2a - \dfrac{2}{a} + 1 = 0}}$

• We should write this as –

$\\ \implies{ \bold{2a - \dfrac{2}{a} = - 1}}$

$\\ \implies{ \bold{a - \dfrac{1}{a} = - \dfrac{1}{2} }}$

• Now take qube on both sides –

$\\ \implies{ \bold{(a - \dfrac{1}{a} )^{3} =( - \dfrac{1}{2} ) ^{3} }}$

• We know that –

$\\ \longrightarrow{ \red{ \boxed{ \bold{(a -b )^{3} = {a}^{ 3} - {b}^{3} - 3ab(a - b ) }}}}$

• So that –

$\\ \implies{ \bold{ {a}^{ 3} - { (\frac{1}{a} )}^{3} - 3a( \frac{1}{a} )(a - \frac{1}{a} ) = - \frac{1}{8} }}$

$\\ \implies{ \bold{ {a}^{ 3} - \frac{1}{ {a}^{3} } - 3( - \frac{1}{2} ) = - \frac{1}{8} }}$

$\\ \implies{ \bold{ {a}^{ 3} - \frac{1}{ {a}^{3} } + \frac{3}{2} = - \frac{1}{8} }}$

$\\ \implies{ \bold{ {a}^{ 3} - \frac{1}{ {a}^{3} } = - \frac{3}{2} - \frac{1}{8} }}$

$\\ \implies{ \bold{ {a}^{ 3} - \frac{1}{ {a}^{3} } = - \frac{13}{8} }}$

• Now add "2" on both sides –

$\\ \implies{ \bold{ {a}^{ 3} - \frac{1}{ {a}^{3} } + 2 = 2 - \frac{13}{8} }}$

$\\ \implies{ \bold{ {a}^{ 3} - \frac{1}{ {a}^{3} } + 2 = \frac{16 - 13}{8} }}$

$\\ \implies{ \red { \boxed{ \bold{ {a}^{ 3} - \frac{1}{ {a}^{3} } + 2 = \frac{3}{8} }}}}$